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## Homework Statement

Answer

Chapter 6, Problem 73 Cutnell & Johnson 7th edition

A 356 N force is pulling an 90.2-kg refrigerator across a horizontal surface. The force acts at an angle of 25.6 ° above the surface. The coefficient of kinetic friction is 0.258, and the refrigerator moves a distance of 9.55 m. Find (a) the work done by the pulling force, and (b) the work done by the kinetic frictional force.

(a)

Number

3066.050223272106 Units J

(b)

Number

-1798.985720775539 Units J

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First :

**sig figs are turned off in my course. The listed answers have to be within 2%.**

**Please no lectures about sig fig. In any other course I obey the rules. ;)**

Incorrect.

(a) Number 3066.050223 Units J (is correct)

(b) Number -1964.181362 Units J (is incorrect)

The 3066.050223J work for the part (a) is correct. Part (b) is the incorrect part I need help with.

Wf = Ff*d*cos 25.6 where Ff = μ*m*g

so 0.258*90.2*-9.8*9.55*cos 180 = -2178 J

0.258*90.2*-9.8*9.55*cos 25.6 = -1964.181362

I've exceeded the 3 tries and can see the correct answer of

**-1798.985720775539**, but have no idea how that was arrived at. The hint given with the problem says to set ΣFy = 0 before considering the Wf . ???????

What am I doing wrong?

## Homework Equations

W= F*D cos 25.6

Wf = Ff*d*cos 25.6 where Ff = μ*m*g

FN = mg = weight of the refrigerator

## The Attempt at a Solution

W done by pulling force

W= F*D cos 25.6

W= 356N*9.55m*cos 25.6 = 3066.050223J

work done by the frictional force is

.258*90.2*9.8*9.55*cos 180 = -2178J

0.258*90.2*-9.8*9.55*cos 25.6 = -1964.181362 is incorrect. The correct answer is -1798.985720775539. ???How?????